JEE Advanced · Mathematics · 28. Area Under Curves
Paragraph:
Consider the polynomial \(f(x)=1+2 x+3 x^2+4 x^3\). Let \(s\) be the sum of all distinct real roots of \(f(x)\) and let \(t=|s|\).Question:
The area bounded by the curve \(y=f(x)\) and the lines \(x=0, y=0\) and \(x=t\), lies in the interval
- A
\(\left(\frac{3}{4}, 3\right)\)
- B
\(\left(\frac{21}{64}, \frac{11}{16}\right)\)
- C
\((9,10)\)
- D
\(\left(0, \frac{21}{64}\right)\)
Answer & Solution
Correct Answer
(A)
\(\left(\frac{3}{4}, 3\right)\)
Step-by-step Solution
Detailed explanation
\(\int_0^{1 / 2} f(x) d x < \int_0^t f(x) d x < \int_0^{3 / 4} f(x) d x\) Now, \(\int f(x) d x\)
\[
\begin{gathered}
=\int\left(1+2 x+3 x^2+4 x^3\right) d x \\
=x+x^2+x^3+x^4 \\
\Rightarrow \quad \int_0^{1 / 2} f(x) d x=\frac{15}{16}>\frac{3}{4} \\
\int_0^{3 / 4} f(x) d x=\frac{530}{256} < 3
\end{gathered}
\]
\[
\begin{gathered}
=\int\left(1+2 x+3 x^2+4 x^3\right) d x \\
=x+x^2+x^3+x^4 \\
\Rightarrow \quad \int_0^{1 / 2} f(x) d x=\frac{15}{16}>\frac{3}{4} \\
\int_0^{3 / 4} f(x) d x=\frac{530}{256} < 3
\end{gathered}
\]
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